If \(x \mathrm{~d} y=y(\mathrm{~d} x+y \mathrm{~d} y), y(1)=1, y(x)>0\), then \(y(-3)\) is

\(x \mathrm{~d} y=y(\mathrm{~d} x+y \mathrm{~d} y)\)
\(\Rightarrow y \mathrm{~d} x=\left(x-y^2\right) \mathrm{d} y \Rightarrow \frac{\mathrm{d} x}{\mathrm{~d} y}+\left(-\frac{1}{y}\right) x=-y\)
\(\therefore \quad\) I.F. \(=\mathrm{e}^{\int-\frac{1}{d y} y^y}=\mathrm{e}^{-\log y}=\frac{1}{y}\)
\(\therefore \quad\) Solution of the given equation is
\(x \cdot \frac{1}{y}=\int-y \cdot \frac{1}{y} \mathrm{~d} y+\mathrm{c}\)
\(\Rightarrow \frac{x}{y}=-y+\mathrm{c}\) ....(i)
Since \(y(1)=1\), i.e., \(y=1\) when \(x=1\)
\(\therefore \quad 1=-1+c \Rightarrow c=2\)
\(\therefore \quad \frac{x}{y}=-y+2 \quad \ldots[\) From (i) \(]\)
Putting \(x=-3\), we get
\(-\frac{3}{y}=-y+2\)
\(\begin{aligned} & \Rightarrow y^2-2 y-3=0 \\ & \Rightarrow(y-3)(y+1)=0\end{aligned}\)
Since \(y(x)>0, y=3\)